Q30 / 50 · Manipal (MET) 2024 · Mathematics

Let $f(x)$ is a polynomial function such that $f(x)+f\left(\frac{1}{x}\right)=f(x) \cdot f\left(\frac{1}{x}\right), x\gt0$ and $f(3)=10$, if $\int f(x) d x=g(x)+c$ and $g(1)=\frac{4}{3}$, then the value of $g(3)$ is

  • A
    10
  • B
    9
  • C
    8
  • Correct answer
    D
    12

Solution

Let, $f(x)=1+x^n$ or $f(x)=1-x^n$ $\because f(3)=10$ So, $f(x)=1+x^2$ Then, $\int f(x) d x=\int\left(1+x^2\right) d x=x+\frac{x^3}{3}+c$ $\begin{array}{ll} \therefore & g(x)=x+\frac{x^3}{3}+c \\ \because & g(1)=\frac{4}{3} \\ \Rightarrow & \frac{4}{3}=1+\frac{1}{3}+c \\ \Rightarrow & C=0 \end{array}$ Therefore, $g(x)=x+\frac{x^3}{3}$ and $g(3)=3+\frac{3^3}{3}=12$

Hence, the correct answer is option: 'D'.

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